connection along a map

Question

I'm interested in connection along a map between manifolds, i.e. the pullback construction. Let $f: M \rightarrow N$ be a map and $(V,\nabla^V) \rightarrow N$ be a vectorbundle with connection. Then via pullback construction one obtains a connection on $f^*V \rightarrow M$, i.e. a mapping $\nabla^f: \Gamma(M) \times \Gamma(f^*V) \rightarrow \Gamma(f^*V)$ satisfying the rules one knows of a connection. Furthermore it's caracterized by

$\nabla^f_X(\sigma \circ f)=\nabla^V_{f_*X}\sigma \in \Gamma(f^*V)$

for $X \in \Gamma(TM), \sigma \in \Gamma(V)$

Now my question is how to see the right hand side of the last equation as a section into the pullback bundle. I want to insert a $p \in M$ and get an element in $V_{f(p)}$. But evaluating $\sigma$ at $p$ is both false and doesn't make sense. I could insert $f(p)$ but since this is not a pointwise construction I cannot... I would be grateful for any comments and hints where to find some references.

Answer 1

The following may be helpful: given a section $\sigma : N \to V$, there exists a unique section $f^* \sigma : M \to f^* V$, the pullback of $\sigma$. (This should follow fairly straightforwardly from whatever definition of pullback bundle you're using.)

The point then is that the pullback connection $\nabla^f$ that you're trying to define should satisfy $$ \nabla^f_X (f^* \sigma) = f^* \big( \nabla^V_{f_* X} \sigma\big) $$ whenever $\sigma : N \to V$ is a section. This is just your equation but written a bit more carefully so that everything really does live in the right spaces.

Answer 2

The usual definition for the pullback bundle is $f^{*}V: =\lbrace (p,v) \in M \,\times V \vert \, \pi(v)=f(p) \rbrace \subset M \times V$ where the $\pi_{f^{*}V}$ is just the projection to the first factor. For $\sigma \in \Gamma(V \stackrel{\pi}{\to} N)$ we define $f^{*}\sigma \in \Gamma\left( f^{*}V \stackrel{\pi_{f^{*}V}}{\to} M\right) $ by $p \mapsto (p, \sigma(f(p)))$. I would write the "naturality condition" for the pullback connection $\nabla^{f}_{X}$, $\nabla^{f}_{X}(f^{*}\sigma)=\nabla_{f_{*}X}\sigma$, explicitly as $(\nabla^{f}_{X}f^{*}\sigma)(p) = (p, \nabla_{(f_{*}X)(p)}\sigma)$. Usually the basepoint of the pullback bundle is ignored (as it is sort of irrelevant), so you write $(f^{*}\sigma)(p)=\sigma(f(p))$ and identify $\Gamma(f^{*}V\to M) \equiv \lbrace \xi: M \to V \,\vert\, \xi \, \text{smooth}, \pi \circ \xi = f\rbrace$.

Remark. I don't think $f^{*}(\nabla_{f_{*}X}\sigma)$ is really meaningful as $\nabla_{f_{*}X}\sigma$ is already evaluated at $f(p)$ since $f_{*}X$ dictates where you may evaluate it.